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Let $\phi\colon[0,1] \to \mathbb R$ be such that $\phi,\phi^\prime,\phi^{\prime\prime}$ are continuous on $[0,1]$, then the following inequality holds:

$$\int_0^1\cos x\frac{x\phi^\prime(x)-\phi(x)+\phi(0)}{x^2}\mathrm dx < \frac32\|\phi^{\prime\prime}\|_\infty$$

I have no idea how to solve this problem, could you help me please?

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What does $||\phi''||_{\infty}$ mean?? – Ramana Venkata Oct 8 '11 at 22:48
It's the uniform norm. If $f\colon D\to\mathbb C$, then $\|f\|_\infty:=\sup \{|f(x)| : x\in D\}$. – SL2 Oct 8 '11 at 22:55
Please don't write \mathrm{cos}\;x. Write \cos x (with a backslash) instead. The backslash both prevents italicization and results in proper spacing. With some operators like \max it also causes formatting conventions to be followed, so that in a "displayed" (as opposed to "inline") setting \max_{x\in A} gives you $\max\limits_{x\in A}$. – Michael Hardy Oct 9 '11 at 2:37
up vote 4 down vote accepted

We have \begin{align*} x\phi'(x)-\phi(x)+\phi(0)&=x\phi'(x)-\int_0^x\phi'(t)dt\\ &=\int_0^x\left(\phi'(x)-\phi'(t)dt\right)\\ &=\int_0^x\int_t^x\phi''(s)dsdt, \end{align*} hence $$\left|\frac{x\phi'(x)-\phi(x)+\phi(0)}{x^2}\right|\leq \frac{\lVert \phi''\rVert}{x^2}\int_0^x\int_t^xdsdt=\frac{\lVert \phi''\rVert}{x^2}\int_0^x(x-t)dt=\frac{\lVert \phi''\rVert}2.$$ Therefore, the integral is convergent and
$$\int_0^1\cos x\frac{x\phi'(x)-\phi(x)+\phi(0)}{x^2}dx\leq\frac{\lVert \phi''\rVert}2 \int_0^1\cos xdx =\frac{\lVert \phi''\rVert}2\sin 1\leq \frac{\lVert \phi''\rVert}2,$$ unless I'm misunderstanding something.

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Use Taylor series twice (I write f for $\phi$ cause I'm lazy):

$f(x) = f(0) + xf'(0) + \frac{x^2}{2} f''(c)$ and $f'(x) = f'(0) + xf''(d)$ where $0 \le c, d \le 1$.

From the second one, $f'(0) = f'(x) - xf''(d)$.

Putting this in the first, $f(x) = f(0) + x(f'(x)-xf''(d)) + \frac{x^2}{2} f''(c)$ so $x f'(x) - f(x) + f(0) = -x^2 f''(d) + \frac{x^2}{2}f''(c)$ or $$\frac{x f'(x) - f(x) + f(0)}{x^2} = -f''(d)+f''(c)/2$$.

Putting this in the integral, since $|\cos| \le 1$ gives the result.

It will be interesting to see how much this agrees with the answer entered while I was entering this

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